Question
Given a hexagon $ABCDEF$ inscribed in a circle with $AB = BC, CD = DE, EF = FA$, show that $\overline{AD}, \overline{BE}$, and $\overline{CF}$ are concurrent.
[asy]
unitsize(2 cm);
pair A, B, C, D, E, F, G;
A = dir(85);
B = dir(45);
C = dir(5);
D = dir(-55);
E = dir(-115);
F = dir(-195);
draw(unitcircle);
draw(A--B--C--D--E--F--A);
draw(A--D);
draw(B--E);
draw(C--F);
label("$A$", A, A);
label("$B$", B, NE);
label("$C$", C, NE);
label("$D$", D, SE);
label("$E$", E, SW);
label("$F$", F, NW);
[/asy]
[asy]
unitsize(2 cm);
pair A, B, C, D, E, F, G;
A = dir(85);
B = dir(45);
C = dir(5);
D = dir(-55);
E = dir(-115);
F = dir(-195);
draw(unitcircle);
draw(A--B--C--D--E--F--A);
draw(A--D);
draw(B--E);
draw(C--F);
label("$A$", A, A);
label("$B$", B, NE);
label("$C$", C, NE);
label("$D$", D, SE);
label("$E$", E, SW);
label("$F$", F, NW);
[/asy]
Answers
oobleck
It should be easy to show that since the three segments are diameters of the hexagon, they all meet in the center.
Dillion
They aren't diameters tho.